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Diffstat (limited to 'source/know/concept/matsubara-greens-function')
| -rw-r--r-- | source/know/concept/matsubara-greens-function/index.md | 103 |
1 files changed, 53 insertions, 50 deletions
diff --git a/source/know/concept/matsubara-greens-function/index.md b/source/know/concept/matsubara-greens-function/index.md index 81dd360..fdcadb3 100644 --- a/source/know/concept/matsubara-greens-function/index.md +++ b/source/know/concept/matsubara-greens-function/index.md @@ -21,10 +21,10 @@ $$\begin{aligned} } \end{aligned}$$ -Where the expectation value $\Expval{}$ is with respect to thermodynamic equilibrium, -and $\mathcal{T}$ is the [time-ordered product](/know/concept/time-ordered-product/) pseudo-operator. -Because the Hamiltonian $\hat{H}$ cannot depend on the imaginary time, -$C_{AB}$ is a function of the difference $\tau \!-\! \tau'$ only: +Where the expectation value $$\Expval{}$$ is with respect to thermodynamic equilibrium, +and $$\mathcal{T}$$ is the [time-ordered product](/know/concept/time-ordered-product/) pseudo-operator. +Because the Hamiltonian $$\hat{H}$$ cannot depend on the imaginary time, +$$C_{AB}$$ is a function of the difference $$\tau \!-\! \tau'$$ only: $$\begin{aligned} C_{AB}(\tau, \tau') @@ -36,9 +36,9 @@ $$\begin{aligned} &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B} \Big) \end{aligned}$$ -For $\tau > \tau'$, we see by expanding in the many-particle eigenstates $\Ket{n}$ -that we need to demand $\hbar \beta > \tau \!-\! \tau'$ to prevent -$C_{AB}$ from diverging for increasing temperatures: +For $$\tau > \tau'$$, we see by expanding in the many-particle eigenstates $$\Ket{n}$$ +that we need to demand $$\hbar \beta > \tau \!-\! \tau'$$ to prevent +$$C_{AB}$$ from diverging for increasing temperatures: $$\begin{aligned} C_{AB}(\tau \!-\! \tau') @@ -48,8 +48,8 @@ $$\begin{aligned} &= - \frac{1}{\hbar Z} \sum_{n} \Matrixel{n}{\hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n} e^{-\beta E_n} e^{(\tau - \tau') E_n / \hbar} \end{aligned}$$ -And likewise, for $\tau < \tau'$, -we must demand that $\tau \!-\! \tau' > -\hbar \beta$ +And likewise, for $$\tau < \tau'$$, +we must demand that $$\tau \!-\! \tau' > -\hbar \beta$$ for the same reason: $$\begin{aligned} @@ -61,14 +61,14 @@ $$\begin{aligned} &= \mp \frac{1}{\hbar Z} \sum_{n} \Matrixel{n}{\hat{B} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n} e^{-\beta E_n} e^{- (\tau - \tau') E_n / \hbar} \end{aligned}$$ -With $-$ for bosons, and $+$ for fermions, -due to the time-ordered product for $\tau > \tau'$. +With $$-$$ for bosons, and $$+$$ for fermions, +due to the time-ordered product for $$\tau > \tau'$$. -On this domain $[-\hbar \beta, \hbar \beta]$, -the Matsubara Green's function $C_{AB}$ +On this domain $$[-\hbar \beta, \hbar \beta]$$, +the Matsubara Green's function $$C_{AB}$$ obeys a useful shift relation: -it is $\hbar \beta$-periodic for bosons, -and $\hbar \beta$-antiperiodic for fermions: +it is $$\hbar \beta$$-periodic for bosons, +and $$\hbar \beta$$-antiperiodic for fermions: $$\begin{aligned} \boxed{ @@ -88,8 +88,8 @@ $$\begin{aligned} <label for="proof-period">Proof</label> <div class="hidden" markdown="1"> <label for="proof-period">Proof.</label> -First $\tau \!-\! \tau' < 0$. -We insert the argument $\tau \!-\! \tau' \!+\! \hbar \beta$, +First $$\tau \!-\! \tau' < 0$$. +We insert the argument $$\tau \!-\! \tau' \!+\! \hbar \beta$$, and use the cyclic property: $$\begin{aligned} @@ -105,8 +105,8 @@ $$\begin{aligned} &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{B}(\tau') \hat{A}(\tau) \Big) \end{aligned}$$ -Since $\tau < \tau'$ by assumption, -we can bring back the time-ordered product $\mathcal{T}$: +Since $$\tau < \tau'$$ by assumption, +we can bring back the time-ordered product $$\mathcal{T}$$: $$\begin{aligned} C_{AB}(\tau \!-\! \tau' \!+\! \hbar \beta) @@ -115,7 +115,7 @@ $$\begin{aligned} &= \pm C_{AB}(\tau \!-\! \tau') \end{aligned}$$ -Moving on to $\tau \!-\! \tau' > 0$, the proof is perfectly analogous: +Moving on to $$\tau \!-\! \tau' > 0$$, the proof is perfectly analogous: $$\begin{aligned} C_{AB}(\tau \!-\! \tau' \!-\! \hbar \beta) @@ -133,16 +133,17 @@ $$\begin{aligned} \\ &= \pm C_{AB}(\tau \!-\! \tau') \end{aligned}$$ + </div> </div> -Due to this limited domain $\tau \in [-\hbar \beta, \hbar \beta]$, +Due to this limited domain $$\tau \in [-\hbar \beta, \hbar \beta]$$, the [Fourier transform](/know/concept/fourier-transform/) -of $C_{AB}(\tau)$ consists of discrete frequencies -$k_n \equiv n \pi / (\hbar \beta)$. +of $$C_{AB}(\tau)$$ consists of discrete frequencies +$$k_n \equiv n \pi / (\hbar \beta)$$. The forward and inverse Fourier transforms -are therefore defined as given below (with $\tau' = 0$). -It is convention to write $C_{AB}(i k_n)$ instead of $C_{AB}(k_n)$: +are therefore defined as given below (with $$\tau' = 0$$). +It is convention to write $$C_{AB}(i k_n)$$ instead of $$C_{AB}(k_n)$$: $$\begin{aligned} \boxed{ @@ -162,8 +163,8 @@ $$\begin{aligned} <div class="hidden" markdown="1"> <label for="proof-FT-def">Proof.</label> We will prove that one is indeed the inverse of the other. -We demand that the inverse FT of the forward FT of $C_{AB}(\tau)$ -is simply $C_{AB}(\tau)$ again: +We demand that the inverse FT of the forward FT of $$C_{AB}(\tau)$$ +is simply $$C_{AB}(\tau)$$ again: $$\begin{aligned} C_{AB}(\tau) @@ -197,11 +198,12 @@ $$\begin{aligned} \\ &= C_{AB}(\tau) \end{aligned}$$ + </div> </div> -Let us now define the **Matsubara frequencies** $\omega_n$ -as a species-dependent subset of $k_n$: +Let us now define the **Matsubara frequencies** $$\omega_n$$ +as a species-dependent subset of $$k_n$$: $$\begin{aligned} \boxed{ @@ -232,7 +234,7 @@ $$\begin{aligned} <div class="hidden" markdown="1"> <label for="proof-FT-alt">Proof.</label> We split the integral, shift its limits, -and use the (anti)periodicity of $C_{AB}$: +and use the (anti)periodicity of $$C_{AB}$$: $$\begin{aligned} C_{AB}(i k_n) @@ -247,9 +249,9 @@ $$\begin{aligned} &= \frac{1}{2} \big( 1 \pm e^{-i k_n \hbar \beta} \big) \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau} \end{aligned}$$ -With $+$ for bosons, and $-$ for fermions. Since $k_n \equiv n \pi / (\hbar \beta)$, -we know $e^{-i k_n \hbar \beta} \in \{-1, 1\}$, -so for bosons all odd $n$ vanish, and for fermions all even $n$, +With $$+$$ for bosons, and $$-$$ for fermions. Since $$k_n \equiv n \pi / (\hbar \beta)$$, +we know $$e^{-i k_n \hbar \beta} \in \{-1, 1\}$$, +so for bosons all odd $$n$$ vanish, and for fermions all even $$n$$, yielding the desired result. For the other case, we simply shift the first integral's limits instead of the seconds': @@ -263,11 +265,12 @@ $$\begin{aligned} \\ &= \frac{1}{2} \big( 1 \pm e^{-i k_n \hbar \beta} \big) \int_{-\hbar \beta}^0 C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau} \end{aligned}$$ + </div> </div> If we actually evaluate this, -we obtain the following form of $C_{AB}$, +we obtain the following form of $$C_{AB}$$, which is almost identical to the [Lehmann representation](/know/concept/lehmann-representation/) of the "ordinary" retarded and advanced Green's functions: @@ -285,8 +288,8 @@ $$\begin{aligned} <label for="proof-Lehmann">Proof</label> <div class="hidden" markdown="1"> <label for="proof-Lehmann">Proof.</label> -For $\tau \!-\! \tau' > 0$, we start by expanding -in the many-particle eigenstates $\Ket{n}$: +For $$\tau \!-\! \tau' > 0$$, we start by expanding +in the many-particle eigenstates $$\Ket{n}$$: $$\begin{aligned} C_{AB}(\tau \!-\! \tau') @@ -300,7 +303,7 @@ $$\begin{aligned} \matrixel{n'}{\hat{B}}{n} e^{(E_n - E_{n'})(\tau - \tau') / \hbar} \end{aligned}$$ -We take the Fourier transform by integrating over $[0, \hbar \beta]$: +We take the Fourier transform by integrating over $$[0, \hbar \beta]$$: $$\begin{aligned} C_{AB}(i \omega_m) @@ -320,8 +323,8 @@ $$\begin{aligned} \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) \end{aligned}$$ -Moving on to $\tau \!-\! \tau' < 0$, -we again expand in the many-particle eigenstates $\Ket{n}$: +Moving on to $$\tau \!-\! \tau' < 0$$, +we again expand in the many-particle eigenstates $$\Ket{n}$$: $$\begin{aligned} C_{AB}(\tau \!-\! \tau') @@ -335,8 +338,8 @@ $$\begin{aligned} \matrixel{n'}{\hat{A}}{n} e^{-(E_n - E_{n'})(\tau - \tau') / \hbar} \end{aligned}$$ -Since $\tau \!-\! \tau' < 0$ this time, -we take the Fourier transform over $[-\hbar \beta, 0]$: +Since $$\tau \!-\! \tau' < 0$$ this time, +we take the Fourier transform over $$[-\hbar \beta, 0]$$: $$\begin{aligned} C_{AB}(i \omega_m) @@ -359,17 +362,17 @@ $$\begin{aligned} \Big( e^{- \beta E_{n'}} \mp e^{-\beta E_n} \Big) \end{aligned}$$ -Where swapping $n$ and $n'$ gives the desired result. +Where swapping $$n$$ and $$n'$$ gives the desired result. </div> </div> -This gives us the primary use of the Matsubara Green's function $C_{AB}$: -calculating the retarded $C_{AB}^R$ and advanced $C_{AB}^A$. -Once we have an expression for Matsubara's $C_{AB}$, -we can recover $C_{AB}^R$ and $C_{AB}^A$ by substituting -$i \omega_m \to \omega \!+\! i \eta$ and $i \omega_m \to \omega \!-\! i \eta$ respectively. +This gives us the primary use of the Matsubara Green's function $$C_{AB}$$: +calculating the retarded $$C_{AB}^R$$ and advanced $$C_{AB}^A$$. +Once we have an expression for Matsubara's $$C_{AB}$$, +we can recover $$C_{AB}^R$$ and $$C_{AB}^A$$ by substituting +$$i \omega_m \to \omega \!+\! i \eta$$ and $$i \omega_m \to \omega \!-\! i \eta$$ respectively. -In general, we can define the **canonical Green's function** $C_{AB}(z)$ +In general, we can define the **canonical Green's function** $$C_{AB}(z)$$ on the complex plane: $$\begin{aligned} @@ -380,8 +383,8 @@ $$\begin{aligned} This is a [holomorphic function](/know/concept/holomorphic-function/), except for poles on the real axis. -It turns out that $C_{AB}(z)$ must have these properties -for the substitution $i \omega_n \to \omega \!\pm\! i \eta$ to be valid. +It turns out that $$C_{AB}(z)$$ must have these properties +for the substitution $$i \omega_n \to \omega \!\pm\! i \eta$$ to be valid. |